Question: Sides $\overline{AH}$ and $\overline{CD}$ of regular octagon $ABCDEFGH$ are extended to meet at point $P$. What is the degree measure of angle $P$?
Answer: The sum of the angle measures of an octagon is $180(8-2) = 1080$ degrees, so each angle of a regular octagon measures $1080^\circ/8=135^\circ$.  Therefore, $\angle BCD= 135^\circ$, which means $\angle BCP = 180^\circ - \angle BCD = 45^\circ$.  Similarly, $\angle PAB = 45^\circ$.  Since $\angle ABC = 135^\circ$, the reflex angle at $B$ that is an interior angle of $ABCP$ has measure $360^\circ - 135^\circ = 225^\circ$.  The interior angles of quadrilateral $ABCP$ must sum to $360^\circ$, so we have  \begin{align*}
\angle P &= 360^\circ - \angle PAB - (\text{reflex }\angle B) - \angle BCP\\
&=360^\circ - 45^\circ - 225^\circ - 45^\circ = \boxed{45^\circ}.
\end{align*} [asy]
unitsize(0.6inch);
pair A,B,C,D,EE,F,G,H,P;

A = rotate(-67.5)*(1,0);
B = rotate(45)*A;
C=rotate(45)*B;
D = rotate(45)*C;
EE = rotate(45)*D;
F = rotate(45)*EE;
G = rotate(45)*F;
H = rotate(45)*G;

P = A + (rotate(-90)*(D-A));
draw (A--B--C--D--EE--F--G--H--A--P--C,linewidth(1));
label("$A$",A,S);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,N);
label("$E$",EE,N);
label("$F$",F,W);
label("$G$",G,W);
label("$H$",H,S);
label("$P$",P,S);
[/asy]

Note that we also could have solved this problem by noticing that $\overline{FC}\parallel\overline{HA}$, so $\angle APD = \angle FCD$.  Since $\overline{CF}\parallel\overline {ED}$, we have $\angle FCD = 180^\circ - \angle CDE = 180^\circ - 135^\circ = 45^\circ$.